Describe the solutions to the following quadratic equation: $2x^{2}+4x+4 = 0$
Explanation: We could use the quadratic formula to solve for the solutions and see what they are, but there's a shortcut... $ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ Think about what the part of the quadratic formula under the radical tells us about the solutions. Substitute the $a$ $b$ , and $c$ coefficients from the quadratic equation: $ \begin{array} && b^2-4ac \\ \\ =& 4^2 - 4 ( 2)(4) \\ \\ =& -16 \end{array} $ Because ${b^2 - 4ac}$ is negative, its square root is imaginary and the quadratic formula reduces to $\dfrac{-b \pm \sqrt{-16}}{2a} $, which means there are two complex solutions.